Question: The circumference of a circle is increasing at a rate of $\dfrac{\pi}{2}$ meters per hour. At a certain instant, the circumference is $12\pi$ meters. What is the rate of change of the area of the circle at that instant (in square meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $3\pi$ (Choice B) B $6$ (Choice C) C $36\pi$ (Choice D) D $\dfrac{\pi}{4}$
Setting up the math Let... $r(t)$ denote the circle's radius at time $t$, $A(t)$ denote the circle's area at time $t$, and $C(t)$ denote the circle's circumference at time $t$. We are given that $C'(t)=\dfrac{\pi}{2}$, We are also given that $C(t_0)=12\pi$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures $C(t)$ and $r(t)$ relate to each other through the formula for the circumference of a circle: $C(t)=2\pi r(t)$ We can differentiate both sides to find an expression for $C'(t)$ : $C'(t)=2\pi r'(t)$ $A(t)$ and $r(t)$ relate to each other through the formula for the area of a circle: $A(t)=\pi[r(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2\pi r(t)r'(t)$ Using the information to solve Let's plug ${C(t_0)}={12\pi}$ into the expression for $C(t_0)$ : $\begin{aligned} {C(t_0)}&=2\pi r(t_0) \\\\ {12\pi}&=2\pi r(t_0) \\\\ {6}&={r(t_0)} \end{aligned}$ Let's plug ${C'(t_0)}={\dfrac{\pi}{2}}$ into the expression for $C'(t_0)$ : $\begin{aligned} {C'(t_0)}&=2\pi r'(t_0) \\\\ {\dfrac{\pi}{2}}&=2\pi r'(t_0) \\\\ C{\dfrac{1}{4}}&=C{r'(t_0)} \end{aligned}$ Now let's plug ${r(t_0)}={6}$ and $C{r'(t_0)}=C{\dfrac{1}{4}}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2\pi{r(t_0)}C{r'(t_0)} \\\\ &=2\pi({6})\left(C{\dfrac{1}{4}}\right) \\\\ &=3\pi \end{aligned}$ In conclusion, the rate of change of the area of the circle at that instant is $3\pi$ square meters per hour. Since the rate of change is positive, we know that the area is increasing.